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### Section 7.6 : Integrals Involving Quadratics

1. Evaluate the integral $$\displaystyle \int{{\frac{7}{{{w^2} + 3w + 3}}\,dw}}$$.

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The first thing to do is to complete the square (we’ll leave it to you to verify the completing the square details) on the quadratic in the denominator.

$\int{{\frac{7}{{{w^2} + 3w + 3}}\,dw}} = \int{{\frac{7}{{{{\left( {w + \frac{3}{2}} \right)}^2} + \frac{3}{4}}}\,dw}}$ Show Step 2

From this we can see that the following substitution should work for us.

$u = w + \frac{3}{2}\hspace{0.5in} \Rightarrow \hspace{0.5in}du = dw$

Doing the substitution gives,

$\int{{\frac{7}{{{w^2} + 3w + 3}}\,dw}} = \int{{\frac{7}{{{u^2} + \frac{3}{4}}}\,du}}$ Show Step 3

This integral can be done with the formula given at the start of this section.

$\int{{\frac{7}{{{w^2} + 3w + 3}}\,dw}} = \frac{{14}}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{{2u}}{{\sqrt 3 }}} \right) + c = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{14}}{{\sqrt 3 }}{{\tan }^{ - 1}}\left( {\frac{{2w + 3}}{{\sqrt 3 }}} \right) + c}}$

Don’t forget to back substitute in for $$u$$!